A study of the mean value of the error term in the mean square formula of the Riemann zeta-function in the critical strip 3/4 ≤ σ < 1
Tome 18, no2 (2006), p. 445-470.
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de Bordeaux 18 (2006), 445-470
A study of the mean value of the error term in
the mean square formula of the Riemann
zeta-function in the critical strip 3/4 ≤ σ < 1
par Yuk-Kam LAU
Résumé. Pour σ dans la bandre critique 1/2 < σ < 1, on note Eσ(T ) le terme d’erreur de la formule asymptotique de
RT
1 |ζ(σ + it)|
2dt (pour T grand). C’est un analogue du terme
d’erreur classique E(T ) (= E1/2(T )). L’étude de E(T ) a une
longue histoire, mais celle de Eσ(T ) est assez récente. En
par-ticulier, lorsque 3/4 < σ < 1, on connaît peu d’informations sur Eσ(T ). Pour en gagner, nous étudions la moyenne
RT
1 Eσ(u) du.
Dans cet article, nous donnons une expression en série de type Atkinson et explorons quelques une des propriétés de la moyenne comme fonction en T .
Abstract. Let Eσ(T ) be the error term in the mean square
formula of the Riemann zeta-function in the critical strip 1/2 < σ < 1. It is an analogue of the classical error term E(T ). The research of E(T ) has a long history but the investigation of Eσ(T ) is quite new. In particular there is only a few
informa-tion known about Eσ(T ) for 3/4 < σ < 1. As an exploration, we
study its mean value RT
1 Eσ(u) du. In this paper, we give it an
Atkinson-type series expansion and explore many of its properties as a function of T .
1. Introduction
Let ζ(s) be the Riemann zeta-function, and let E(T ) =
Z T
0
|ζ(1/2 + it)|2dt − T (log T
2π + 2γ − 1)
denote the error term in the mean-square formula for ζ(s) (on the critical line). The behaviour of E(T ) is interesting and many papers are devoted
to study this function. Analogously, it is defined for 1/2 < σ < 1, Eσ(T ) = Z T 0 |ζ(σ + it)|2dt −ζ(2σ)T + (2π)2σ−1ζ(2 − 2σ) 2 − 2σ T 2−2σ.
The behaviour of Eσ(T ) is very interesting too, and in fact, more delicate
analysis is required to explore its properties such as the Atkinson-type series expansion and mean square formula, see ([17]-[20]). Excellent surveys are given in [11] and [18].
In the critical strip 1/2 < σ < 1, our knowledge of Eσ(T ) is not ‘uniform’,
for example, an asymptotic formula for the mean square is available for 1/2 < σ ≤ 3/4 but not for the other part. In fact, not much is known for the case 3/4 < σ < 1, except perhaps some upper bound estimates and
(1.1) T
Z T
1
Eσ(t)2dt T (3/4 < σ < 1).
(See [7], [20] and [14].) To furnish this part, we look at the mean value
RT
1 Eσ(u) du. The mean values of E(T ) and Eσ(T ) (1/2 < σ < 3/4) are
respectively studied in [2] and [6], each of which gives an Atkinson-type expansion. Correspondingly, we prove an analogous formula with a good error term in the case 3/4 ≤ σ < 1. Actually, the tight lower bound in (1.1) is shown in [14] based on this formula. The proof of the asymptotic formula relies on the argument of [2] and uses the tools available in [2] and [19]. But there is a difficulty which we need to get around. In [2], Hafner and Ivić used a result of Jutila [9] on transformation of Dirichlet Polynomials, which depends on the formula
P0 a≤n≤bd(n)f (n) = Z b a (log x + 2γ)f (x) dx + ∞ X n=1 d(n) Z b a f (x)α(nx) dx, where α(x) = 4K0(4π √ x) − 2πY0(4π √
x) is a combination of the Bessel functions K0 and Y0. It is not available in our case but this can be avoided
by using an idea in [19].
In addition, we shall regard the mean value as a function of T and study its behaviour; more precisely, we consider
(1.2) Gσ(T ) =
Z T
1
(Eσ(t) + 2πζ(2σ − 1)) dt.
(The remark below Corollary 1 explains the inclusion of 2πζ(2σ − 1)T .) Unlike the case 1/2 ≤ σ < 3/4, the function Gσ(T ) is now more
fluctu-ating. Nevertheless we can still explore many interesting properties, in-cluding some power moments, Ω±-results, gaps between sign-changes and
limiting distribution functions, by using the tools in [19], [23], [4], [3], [1] and [13]. Particularly, we can determine the exact order of magnitude of the gaps of sign-changes (see Theorems 5 and 6). The limiting distribution
function is not computed in the case 1/2 ≤ σ < 3/4, perhaps because it is less interesting in the sense that the exact order of magnitude of Gσ(t)
(1/2 ≤ σ < 3/4) is known; therefore, the limiting distribution is ‘compactly supported’. Here, a limiting distribution P (u) is said to be compactly sup-ported if P (u) = 0 for all u ≤ a and P (u) = 1 for all u ≥ b, for some constants a < b. (Note that a distribution function is non-decreasing.) However, in our case the distribution never vanishes (i.e. never equal to 0 or 1), and we evaluate the rate of decay.
2. Statement of results
Throughout the paper, we assume 3/4 ≤ σ < 1 to be fixed and use c, c0 and c00 to denote some constants which may differ at each occurrence. The implied constants in - or O-symbols and the unspecified positive constants ci (i = 1, 2, . . .) may depend on σ.
Let σa(n) =Pd|nda and arsinh x = log(x +
√ x2+ 1). We define Σ1(t, X) = √ 2 t 2π 5/4−σ X n≤X (−1)nσ1−2σ(n) n7/4−σ e2(t, n) sin f (t, n), Σ2(t, X) = 2 t 2π 1/2−σ X n≤B(t,√X) σ1−2σ(n) n1−σ log t 2πn −2 sin g(t, n), where e2(t, n) = 1 +πn 2t −1/4 r2t πn arsinh rπn 2t !−2 , f (t, n) = 2t arsinh r πn 2t + (2πnt + π 2n2)1/2−π 4, g(t, n) = t log t 2πn− t + π 4, B(t, √ X) = t 2π + X 2 − √ X t 2π+ X 4 1/2 = t 2π + X 4 1/2 − √ X 2 2 .
Theorem 1. Let σ ∈ [3/4, 1), T ≥ 1 and N T . We have
Z T
1
Eσ(t) dt = −2πζ(2σ − 1)T + Σ1(T, N ) − Σ2(T, N ) + O(log2T ).
The next result follows with the trivial bounds on Σ1(T, N ) and Σ2(T, N ). Corollary 1. For all T ≥ 3, we have
Z T
1
Eσ(t) dt = −2πζ(2σ − 1)T + O(
√
and
Z T
1
E3/4(t) dt = −2πζ(1/2)T + O(√T log T ).
Remark. It suggests that Eσ(t) is a superimposition of the constant
−2πζ(2σ − 1) and an oscillatory function, say, Eσ∗(t). (Indeed this view-point had appeared in [18].)
Define Gσ(t) =
Z t
1
Eσ∗(t) dt, which is (1.2). Then,
Gσ(t) t1/2 (3/4 < σ < 1) and G3/4(t) t1/2log T.
Integrating termwisely with partial integrations, one gets (2.1)
Z 2T
T
Gσ(t) dt = o(T1+(5/4−σ)) (3/4 ≤ σ < 1).
In addition, we have the following higher power moments.
Theorem 2. Let σ ∈ [3/4, 1) and T ≥ 1. We have
(1) Z 2T T Gσ(t)2dt = B(σ) Z 2T T (t/(2π))5/2−2σdt + O(T3−2σ), (2) Z 2T T Gσ(t)3dt = −C(σ) Z 2T T (t/(2π))15/4−3σdt + O(T(13−8σ)/3),
where B(σ) and C(σ) are defined by
B(σ) = ∞ X n=1 σ1−2σ(n)2n2σ−7/2 = ζ(7/2 − 2σ)ζ(3/2 + 2σ)ζ(5/2)2ζ(5)−1, C(σ) = 3 2 ∞ X s=1 µ(s)2 s21/4−3σ ∞ X a,b=1 σ1−2σ(sa2) a7/2−2σ σ1−2σ(sb2) b7/2−2σ σ1−2σ(s(a + b)2) (a + b)7/2−2σ .
(3) for any real k ∈ [0, A0) and any odd integer l ∈ [0, A0) where A0 =
(σ − 3/4)−1, Z T 1 |Gσ(t)|kdt ∼ αk(σ)T1+k(5/4−σ) and Z T 1 Gσ(t)ldt ∼ βl(σ)T1+l(5/4−σ).
for some constants αk(σ) > 0 and βl(σ) depending on σ. (A0 de-notes ∞ when σ = 3/4.)
Remark. We have no information about the value of βl(σ), which may be
positive, negative or even zero. (See [16] for possible peculiar properties of series of this type.)
It is expected that Gσ(t) is oscillatory and its order of magnitude of Gσ(t)
is about t5/4−σ. (2.1) shows a big cancellation between the positive and negative parts, but Theorem 2 (2) suggests that it skews towards negative. This phenomenon also appears in the case 1/2 ≤ σ < 3/4. Now, we look at its distribution of values from the statistical viewpoint.
Theorem 3. For 3/4 ≤ σ < 1, the limiting distribution Dσ(u) of the function tσ−5/4Gσ(t) exists, and is equal to the distribution of the random series η =P∞ n=1an(tn) where an(t) = √ 2 µ(n) 2 n7/4−σ ∞ X r=1 (−1)nrσ1−2σ(nr 2) r7/2−2σ sin(2πrt − π/4)
and tn’s are independent random variables uniformly distributed on [0, 1]. Define tail(Dσ(u)) = 1 − Dσ(u) for u ≥ 0 and Dσ(u) for u < 0. Then
(2.2) exp(−c1exp(|u|)) tail(D3/4(u)) exp(−c2exp(|u|)), exp(−c3|u|4/(4σ−3)) tail(Dσ(u)) exp(−c4|u|4/(4σ−3)) for 3/4 < σ < 1.
Remark. Dσ(u) is non-symmetric and skews towards the negative side be-cause of Theorem 2 (2). Again it is true for 1/2 ≤ σ < 3/4. But in the case 1/2 ≤ σ < 3/4, the closure of the set {u ∈ R : 0 < Dσ(u) < 1} is compact
and it differs from our case.
To investigate the oscillatory nature, we consider the extreme values of Gσ(t) and the frequency of occurrence of large values. These are revealed
in the following three results.
Theorem 4. We have
G3/4(T ) = Ω−(
√
T log log T ) and G3/4(T ) = Ω+(
√
T log log log T ).
For 3/4 < σ < 1, Gσ(T ) = Ω−(T5/4−σ(log T )σ−3/4) and Gσ(T ) = Ω+ T5/4−σexp c5 (log log T )σ−3/4 (log log log T )7/4−σ
Theorem 5. For σ ∈ [3/4, 1) and for every sufficiently large T , there exist
t1, t2∈ [T, T + c6
√
T ] such that Gσ(t1) ≥ c7t5/4−σ1 and Gσ(t2) ≤ −c7t5/4−σ2 . In particular, Gσ(t) has (at least) one sign change in every interval of the form [T, T + c8√T ].
Theorem 6. Let σ ∈ [3/4, 1) and δ > 0 be a fixed small number. Then for all sufficiently large T ≥ T0(δ), there are two sets S+ and S− of disjoint intervals in [T, 2T ] such that
1. every interval in S± is of length c9δ√T , 2. the cardinality of S±≥ c10δ4(1−σ)
√ T ,
3. ±Gσ(t) ≥ (c11− δ5/2−2σ)t5/4−σ for all t ∈ I with I ∈ S± respec-tively.
Remark. Theorems 5 and 6 determine the order of magnitude of the gaps
between sign-changes.
3. Series representation
This section is to prove Theorem 1 and we need two lemmas, which come from [2, Lemma 3] and [19, Lemma 1] with [22] respectively.
Lemma 3.1. Let α, β, γ, a, b, k, T be real numbers such that α, β, γ are positive and bounded, α 6= 1, 0 < a < 1/2, a < T /(8πk), b ≥ T , k ≥ 1,
T ≥ 1, U (t) = t 2πk + 1 4 1/2 , V (t) = 2arsinh s πk 2t, Lk(t) = (2ki √ π)−1t1/2V (t)−γ−1U (t)−1/2 U (t) −1 2 −α U (t) +1 2 −β × exp
itV (t) + 2πikU (t) − πik +πi 4 , and J (T ) = Z 2T T Z b a y−α(1 + y)−β log1 + y y −γ
× exp(it log(1 + 1/y) + 2πiky) dy dt.
Then uniformly for |α − 1| ≥ , 1 ≤ k ≤ T + 1, we have
J (T ) = Lk(2T ) − Lk(T ) + O(a1−α) + O(T k−1bγ−α−β)
+ O((T /k)(γ+1−α−β)/2T−1/4k−5/4).
In the case −k in place of k, the result holds without Lk(2T ) − Lk(T ) for the corresponding integral.
Lemma 3.2. Let ∆1−2σ(t) =P0n≤tσ1−2σ(n) − ζ(2σ)t + ζ(2 − 2σ) 2 − 2σ t 2−2σ− 1 2ζ(2σ − 1)
where the sumP0
n≤t counts half of the last term only when t is an integer. Define ˜∆1−2σ(ξ) = R0ξ∆1−2σ(t) dt − ζ(2σ − 2)/12. Assuming 3/4 ≤ σ < 1, we have for ξ ≥ 1, ˜ ∆1−2σ(ξ) = C1ξ5/4−σ ∞ X n=1 σ1−2σ(n)nσ−7/4cos(4π p nξ + π/4) + C2ξ3/4−σ ∞ X n=1 σ1−2σ(n)nσ−9/4cos(4π p nξ − π/4) + O(ξ1/4−σ)
where the two infinite series on the right-hand side are uniformly convergent on any finite closed subinterval in (0, ∞), and the values of the constants are C1 = −1/(2 √ 2π2), C2 = (5 − 4σ)(7 − 4σ)/(64 √ 2π3). In addition, we have for 3/4 ≤ σ < 1, ∆1−2σ(v) v1−σ, Z x 1 ∆1−2σ(v)2dv x log x, ˜ ∆1−2σ(ξ) ξrlog ξ, Z x 1 ˜ ∆1−2σ(v)2dv x7/2−2σ where 0 < r = −(4σ2− 7σ + 2)/(4σ − 1) ≤ 1/2.
Proof of Theorem 1. From [17, (3.4)] and [20, (3.1)], we have
Z t −t |ζ(σ + iu)|2du = 2ζ(2σ)t + 2ζ(2σ − 1)Γ(2σ − 1)sin(πσ) 1 − σ t 2−2σ − 2i Z σ+it σ−it g(u, 2σ − u) du + O(min(1, |t|−2σ)). (Note that the value of c3 in [20, (3.1)] is zero.) Hence, we have
Eσ(t) = −i Z σ+it σ−it g(u, 2σ − u) du + O(min(1, t−2σ)). Define (3.1) h(u, ξ) = 2 Z ∞ 0
Assume AT ≤ X ≤ T and X is not an integer where 0 < A < 1 is a constant. Then, following [19, p.364-365], we define
G1(t) = X n≤X σ1−2σ(n) Z σ+it σ−it h(u, n) du, G2(t) = ∆1−2σ(X) Z σ+it σ−it h(u, X) du, G3(t) = Z σ+it σ−it Z ∞ X (ζ(2σ) + ζ(2 − 2σ)ξ1−2σ)h(u, ξ) dξ du, (3.2) G∗4(t) = ˜∆1−2σ(X) Z σ+it σ−it ∂h ∂ξ(u, X) du, G∗∗4 (t) = Z ∞ X ˜ ∆1−2σ(ξ) Z σ+it σ−it ∂2h ∂ξ2(u, ξ) du dξ. Then, we have Z 2T T Eσ(t) dt = −i Z 2T T G1(t) dt + i Z 2T T G2(t) dt − i Z 2T T G3(t) dt − i Z 2T T G∗4(t) dt − i Z 2T T G∗∗4 (t) dt + O(1) (3.3) 1) Evaluation of R2T T G1(t) dt. By Lemma 3.1 with γ = 1, α = β = σ, we have from (3.1), Z 2T T Z σ+it σ−it h(u, n) du dt = 4i Z 2T T Z ∞ 0
(y(1 + y))−σ(log(1 + 1/y))−1 × sin(t log((1 + y)/y)) cos(2πny) dy dt = 2i Im
Z 2T
T
Z ∞
0
(y(1 + y))−σ(log(1 + 1/y))−1 ×nexp(i(t log((1 + y)/y) + 2πny)) + exp(i(t log((1 + y)/y) − 2πny))ody dt = 2i Im (Ln(2T ) − Ln(T )) + O(T3/4−σnσ−9/4)
Noting that Ln(t) = (i
√
we get with (3.2) that Z 2T T G1(t) dt = √ 2i t 2π 5/4−σ X n≤X (−1)nσ1−2σ(n) n7/4−σ e2(t, n) sin f (t, n) 2T T + O(T3/4−σ). (3.4) 2) Evaluation of R2T
T G2(t) dt. The treatment is similar to G1. From (3.2)
and Lemma 3.1,
Z 2T
T
Z σ+it
σ−it
h(u, X) du dt = 2i Im(LX(2T ) − LX(T )) + O(T3/4−σXσ−9/4).
Since LX(t) t5/4−σXσ−7/4 T−1/2 for t = T or 2T , we have
(3.5) Z 2T T G2(t) dt ∆1−2σ(X)T−1/2 T1/2−σ. 3) Evaluation ofR2T T G3(t) dt. Using [17, (4.6)], we have G3(t) = −2iπ−1(ζ(2σ) + ζ(2 − 2σ)X1−2σ) Z ∞ 0 y−σ−1(1 + y)−σ(log(1 + 1/y))−1 × sin(2πXy) sin(t log(1 + 1/y)) dy + (1 − 2σ)π−1ζ(2 − 2σ)X1−2σ Z ∞ 0 y−1(1 + y)1−2σsin(2πXy) × Z σ+it σ−it (u + 1 − 2σ)−1 1 + y y u du dy. Direct computation shows that for y > 0,
Z σ+it σ−it (u + 1 − 2σ)−1(1 + 1/y)udu = 2πi 1 + y y 2σ−1 + Z σ+it −∞+it + Z −∞−it σ−it (1 + 1/y)u(u + 1 − 2σ)−1du. Then, we have Z 2T T G3(t) dt = 2i(1 − 2σ)ζ(2 − 2σ)T X1−2σI1 − 2iπ−1(ζ(2σ) + ζ(2 − 2σ)X1−2σ)I2 + π−1(1 − 2σ)ζ(2 − 2σ)X1−2σI3 (3.6)
where I1 = Z ∞ 0 y−2σsin(2πXy) dy I2 = Z 2T T Z ∞ 0 y−1−σ(1 + y)−σ(log(1 + 1/y))−1 × sin(2πXy) sin(t log(1 + 1/y)) dy dt, I3 = Z 2T T Z ∞ 0 y−1(1 + y)1−2σsin(2πXy) × Z σ+it −∞+it + Z −∞−it σ−it (1 + 1/y)u(u + 1 − 2σ)−1du dy dt.
Then, I1= 22σ−2π2σX2σ−1/(Γ(2σ) sin(πσ)) which is the main contribution.
Interchanging the integrals, we have
I2 = −
Z ∞
0
y−1−σ(1 + y)−σ log(1 + 1/y)−2
× sin(2πXy) cos(t log(1 + 1/y))
t=2T t=T dy.
We split the integral into two parts Rc
0 +
R∞
c for some large constant
c > 0. Expressing the product sin(· · · ) cos(· · · ) as a combination of exp(i(t log(1 + 1/y) ± 2πXy)), since (d/dy)(t log(1 + 1/y) ± 2πXy) = ±2πX − t/(y(1 + y)) X for y ≥ c (recall t = T or 2T ), the integral R∞
c is
X−1 by the first derivative test. Applying the mean value theorem for
integrals, we have Z c 0 Z c00 c0 y
−1−σ(1 + y)−1sin(2πXy) cos(t log(1 + 1/y)) dy
.
Integration by parts yields that the last integralRc00
c0 equals
t−1y−σsin(2πXy) sin(t log(1 + 1/y)) c00 c0 − Z c00 c0 O(y −σ−1| sin(2πXy)| + y−σX) dy 1. (3.7)
Hence I2 1. For I3, the extra integration over t is in fact not necessary to
yield our bound. Thus, we write I3=RT2T(I31+ I32) dt, separated according
out I31 only. Using integration by parts over u,
I31=
Z ∞
0
y−1(1 + y)1−2σ(log(1 + 1/y))−1sin(2πXy) exp(it log(1 + 1/y)) × ( (1 + 1/y)α α + 1 − 2σ + it α=σ α=−∞ + Z σ −∞(1 + 1/y) α dα (α + 1 − 2σ + it)2 ) dy. Then we consider Z ∞ 0
y−1(1+y)1−2σ(log(1+1/y))−1sin(2πXy) exp(it log(1+1/y))(1+1/y)αdy. Again, we split the integral into Rc
0+ R∞ c . Then R∞ c X −1. If α ≤ −2, thenRc
0 1 trivially; otherwise, we have (see (3.7))
Z c 0 Z c00 c0 y
−1−α(1 + y)−1sin(2πXy) exp(it log(1 + 1/y)) dy
1. Therefore, I31 T−1and so I3 1. Putting these estimates into (3.6), we
get Z 2T T G3(t) dt = i22σ−1π2σ (1 − 2σ)ζ(2 − 2σ) Γ(2σ) sin(πσ) T + O(1) = −2πiζ(2σ − 1)T + O(1). (3.8) 4) Evaluation ofR2T T G ∗
4(t) dt. From [19, Section 4], we obtain
Z 2T
T
G∗4(t) dt = 4i ˜∆1−2σ(X)((2σ − 1)I1+ I2− σI3− I4)
where by Lemma 3.1, (recall LX(t) T−1/2 X−1/2 for t = T or 2T )
I1 = X2σ−2
Z 2T
T
Z ∞
0
cos(2πy) sin(t log(1 + X/y)) yσ(X + y)σlog(1 + X/y) dy dt = X−1 Z 2T T Z ∞ 0
cos(2πXy) sin(t log(1 + 1/y))
yσ(1 + y)σlog(1 + 1/y) dy dt X −3/2 I2 = X2σ−1 Z 2T T t Z ∞ 0
cos(2πy) cos(t log(1 + X/y)) yσ(X + y)σ+1log(1 + X/y) dy dt X−1T sup T ≤T1≤T2≤2T Z T2 T1 Z ∞ 0
cos(2πXy) cos(t log(1 + 1/y)) yσ(1 + y)σ+1log(1 + 1/y) dy dt X−1/2
and similarly I3, I4 X−3/2. With Lemma 3.2,
(3.9)
Z 2T
T
5) Evaluation ofR2T
T G
∗∗
4 (t) dt. [19, (3.6) and Section 5] gives
(3.10)
Z 2T
T
G∗∗4 (t) dt = −4iI1+ 4iI2+ 4iI3.
I1, I2 and I3 are defined as follows: write
(3.11) w(ξ, y) = ˜∆1−2σ(ξ)ξ−2y−σ(1 + y)−σ−2(log(1 + 1/y))−1cos(2πξy),
then I1 = Z ∞ X Z 2T T t2 Z ∞ 0
w(ξ, y) sin(t log(1 + 1/y)) dy dt dξ I2 = Z ∞ X Z 2T T t Z ∞ 0
w(ξ, y)H1(y) cos(t log(1 + 1/y)) dy dt dξ
I3 = Z ∞ X Z 2T T Z ∞ 0
w(ξ, y)H0(y) sin(t log(1 + 1/y)) dy dt dξ
where H0(y) and H1(y) are linear combinations of yµ(log(1 + 1/y))−ν with
µ + ν ≤ 2 and µ + ν ≤ 1 respectively. (Remark: It is stated in [19] µ + ν ≤ 2 only for both H0(y) and H1(y).)
When ξ ≥ X T t and µ + ν ≤ 2, we have
Z 2T
T
Z ∞
0
exp(it log(1 + 1/y)) cos(2πξy)
yσ−µ(1 + y)σ+2(log(1 + 1/y))ν+1dy dt 1,
(3.12)
Z ∞
0
exp(it log(1 + 1/y)) cos(2πξy)
yσ−µ(1 + y)σ+2(log(1 + 1/y))ν+1 dy T −1/2
. (3.13)
The estimate (3.13) can be seen from [19, p.368]. To see (3.12), we split the inner integral into Rc
0 +
R∞
c . First derivative test gives
R∞
c ξ−1. For
Rc
0,
we integrate over t first and plainly Rc
0
R2T
T 1.
Using (3.12) and Lemma 3.2, we have I3 RX∞∆˜1−2σ(ξ)ξ−2dξ
T1/4−σ. Applying integration by parts to the t-integral, we find that I2
T3/4−σ with (3.12) and (3.13). (Here we have used µ + ν ≤ 1 for H1(y).)
Since
Z 2T
T
t2sin(t log(1 + 1/y)) dt = −t2(log(1 + 1/y))−1cos(t log(1 + 1/y))
2T T
+ 2t(log(1 + 1/y))−2sin(t log(1 + 1/y)) 2T T − 2(log(1 + 1/y))−2 Z 2T T
sin(t log(1 + 1/y)) dt, the last two terms contribute T3/4−σ and T1/4−σ in I1 respectively by using
[5, Lemma 15.1]) and (3.11) Z 2T T G∗∗4 (t) dt = 4it2 Z ∞ X Z ∞ 0
w(ξ, y)(log(1 + 1/y))−1cos(t log(1 + 1/y)) dy dξ
t=2T t=T + O(T3/4−σ) = iπ−1/2t5/2 Z ∞ X ˜ ∆1−2σ(ξ) cos(tV + 2πξU − πξ + π/4) ξ3V2U1/2(U − 1/2)σ(U + 1/2)σ+2 dξ 2T T (3.14) + O(T3/4−σ)
where U and V are defined as in Lemma 3.1 with k replaced by ξ. Applying the argument in [19, Section 6] to (3.14), we get
Z 2T T G∗∗4 (t) dt = −2i t 2π 1/2−σ X n≤B(t,√X) σ1−2σ(n) n1−σ log t 2πn −2 sin g(t, n) t=2T t=T + O(log T ). (3.15)
(Remark: The σ in [19, Lemma 4] should be omitted, as mentioned in [18].) Inserting (3.4), (3.5), (3.8), (3.9), (3.15) into (3.3), we obtain
Z 2T T Eσ(t) dt = −2πζ(2σ − 1)T + Σ1(t, X) 2T T − Σ2(t, X) 2T T + O(log T ). (3.16)
6) Transformation of Dirichlet Polynomial. Let X1, X2 T (both are
not integers) and denote B1 = B(T,
√ X1) and B2 = B(T, √ X2). Assume X1 < X2. Write F (x) = xσ−1 log T 2πx −2 exp(i(T log T 2πx + 2πx − T + π 4)), then we have X B(T,√X2)<n≤B(T,√X1) σ1−2σ(n)nσ−1(log(T /(2πn)))−2sin g(T, n) = Im X B2<n≤B1 σ1−2σ(n)F (n). (3.17)
Stieltjes integration gives X B2<n≤B1 σ1−2σ(n)F (n) = Z B1 B2 F (t)(ζ(2σ) + ζ(2 − 2σ)t1−2σ) dt + ∆1−2σ(t)F (t) B1 B2 − Z B1 B2 ∆1−2σ(t)F0(t) dt = I1+ I2− I3, say. (3.18)
Now, since (d/dt)(g(T, t) + 2πt) = 2π − T /t < −c when B2 < t < B1, we
have I1 = Z B1 B2 (ζ(2σ) + ζ(2 − 2σ)t1−2σ) × tσ−1(log(T /(2πt)))−2exp(i(g(T, t) + 2πt)) dt Tσ−1.
By Lemma 3.2, I2 1. Direct computation gives
F0(t) = i(2π−T t)t σ−1log T 2πt −2 exp(i(T log T 2πt+2πt−T + π 4))+O(t σ−2) where B2 ≤ t ≤ B1. As RB2B1|∆1−2σ(t)|tσ−2dt Tσ−1 √ log T , we have by (3.18) that X B2<n≤B1 σ1−2σ(n)F (n)
= −i exp(i(T log T
2π− T + π 4)) Z B1 B2 ∆1−2σ(t)(2π − T t)t σ−1 × log T 2πt −2
exp(i(2πt − T log t)) dt + O(1). (3.19)
The integralRB2
B1 in (3.19) is, after by parts,
˜ ∆1−2σ(t)(2π − T t)t σ−1log T 2πt −2 exp(i(2πt − T log t)) B2 B1 − Z B1 B2 ˜ ∆1−2σ(t) d dt ( (2π −T t)t σ−1log T 2πt −2 exp(i(2πt − T log t)) ) dt
The first term is Tσ−1/2log T by Lemma 3.2. Besides, computing directly
shows that for B2≤ t ≤ B1,
d
dt{· · · } = i(2πt − T )
2tσ−3log T
2πt
−2
exp(i(2πt − T log t)) + O(tσ−2).
Treating the O-term with Lemma 3.2, (3.19) becomes
X B2<n≤B1 σ1−2σ(n)F (n) = − exp(i(T log T 2π − T + π 4)) Z B1 B2 ˜ ∆1−2σ(t)(2πt − T )2tσ−3 × log T 2πt −2
exp(i(2πt − T log t)) dt + O(Tσ−1/2log T ). (3.20)
Inserting the Voronoi-type series of ˜∆1−2σ(t) (see Lemma 3.2) into (3.20),
we get X B2<n≤B1 σ1−2σ(n)F (n) = − exp(i(T log T 2π − T + π 4)) × ( C1 ∞ X n=1 σ1−2σ(n) n7/4−σ J1(n) + C2 ∞ X n=1 σ1−2σ(n) n9/4−σ J2(n) ) + O(Tσ−1/2log T ) (3.21) where J1(n) = Z B1 B2 (2πt − T )2t−7/4log T 2πt −2 exp(i(2πt − T log t)) × cos(4π√nt + π 4) dt J2(n) = Z B1 B2 (2πt − T )2t−9/4log T 2πt −2 exp(i(2πt − T log t)) × cos(4π√nt − π 4) dt.
Applying the first derivative test or bounding trivially, we have J2(n)
n ≥ c0T . Thus, the second sum in (3.21) is T −1/4 X n≤cT +T3/4 X cT <n<c0T σ1−2σ(n)nσ−9/4 + T1/4 X n≥c0T σ1−2σ(n)nσ−11/4 Tσ−1/2. (3.22)
After a change of variable t = x2,
J1(n) = Z √ B1 √ B2 (2πx2− T )2x−5/2 log T 2πx2 −2
×nexp i(2πx2− 2T log x + 4π√nx +π 4)
+ exp i(2πx2− 2T log x − 4π√nx −π 4)
o
dx.
Then we use [5, Theorem 2.2], with f (x) = x2− π−1T log x, Φ(x) = x3/2, F (x) = T , µ(x) = x/2 and k = ±2√n. Thus, J1(n) = δn2π2 T 2π 3/4 e2(T, n) exp(i(f (T, n) − T log T 2π+ T − πn + 3π 4 )) + O δnT−1/4+ O T3/4exp(−c √ nT − cT ) + O T3/4min(1, |pX1± √ n|−1) + O T3/4min(1, |pX2± √ n|−1)
where δn = 1 if B2 < x0 < B1 and k > 0, or δn = 0 otherwise. (x0 =
p
T /(2π) + n/4 −√n/2 is the saddle point.) Note that B2 < x0 < B1 is
equivalent to X1 < n < X2. Thus, for the first term in (3.21), we have
−C1exp i(T log
T 2π − T + π 4) ∞ X n=1 σ1−2σ(n) n7/4−σ J1(n) = √1 2 T 2π 3/4 X X1<n<X2 σ1−2σ(n) n7/4−σ e2(T, n) exp(i(f (T, n) − πn + π)) + O(Tσ−1/2log T )
Together with (3.22), (3.21) and (3.17), we obtain X B(T,√X2)<n≤B(T,√X1) σ1−2σ(n)nσ−1(log T /(2πn))−2sin g(T, n) = −√1 2 T 2π 3/4 X X1<n<X2 (−1)nσ1−2σ(n) n7/4−σ e2(T, n) sin f (T, n) + O(Tσ−1/2log T ). (3.23)
We can complete our proof now. Taking X = [T ] − 1/2 in (3.16), we have Σi(t, X) − Σi(t, T ) log T for i = 1, 2 and t = T, 2T ; hence
Z 2T T Eσ(t) dt = −2πζ(2σ − 1)T + Σ1(t, t) 2T T − Σ2(t, t) 2T T − Σ1(2T, 2T ) − Σ(2T, T )+ Σ2(2T, T ) − Σ2(2T, 2T ) + O(log T ).
Choosing X1 and X2 in (3.23) to be half-integers closest to T and 2T
respectively, then (Σ1(2T, 2T ) − Σ(2T, T )) + (Σ2(2T, T ) − Σ2(2T, 2T ))
log T . Hence,
Z T
1
Eσ(t) dt = −2πζ(2σ − 1)T + Σ1(T, T ) − Σ2(T, T ) + O(log2T ).
The extra log T in the O-term comes from the number of dyadic intervals. Suppose N T . We apply (3.23) again with X1 = [N ] + 1/2 and X2 =
[T ] + 1/2 to yield our theorem.
4. The second and third power moments
The proof of the second moment is quite standard, see [19], [21] or [5] for example. Part (1) of Theorem 2 follows from that for N T ,
Z 2T T Σ2(t, N )2dt T, Z 2T T Σ1(t, N )Σ2(t, N ) dt T log T, Z 2T T Σ1(t, N )2dt = B(σ) Z 2T T t 2π 5/2−2σ dt + O(T3−2σ). Moreover, one can show
Lemma 4.1. Define ΣM,M0(t) = Σ1,M0(t) − Σ1,M(t) for 1 ≤ M ≤ M0 T . Then, we have
Z 2T
T
ΣM,M0(t)2dt T7/2−2σM2σ−5/2.
The next result is of its own interest and will be used in the proof of the third moment.
Proposition 4.1. Let 0 ≤ A < (σ − 3/4)−1. Then, we have
Z 2T
T
|Gσ(t)|Adt T1+A(5/4−σ).
Proof. The case 0 ≤ A ≤ 2 is proved by Hölder’s inequality and part (1) of
Theorem 2. Consider the situation 2 < A < (σ − 3/4)−1. Then, for T ≤ t ≤ 2T and N T , we have Σ2(t, N ) T1/2 and hence RT2T|Σ2(t, N )|Adt
TA/2. We take N = 2R− 1 T and write M = 2r. Then Σ
1(t, N ) ≤
P
r≤R|ΣM,2M(t)|. By Hölder’s inequality, we have
|Σ1(t, N )|A X r≤R αAr|ΣM,2M(t)|A X r≤R α−A/(A−1)r A−1 .
Taking αr = M(1−A(σ−3/4))/(2A) with the trivial bound ΣM,2M(t)
T5/4−σMσ−3/4, we have Z 2T T |Σ1(t, N )|Adt AT(5/4−σ)(A−2) X r≤R αArM(σ−3/4)(A−2) Z 2T T ΣM,2M(t)2dt AT1+A(5/4−σ) (4.1) by Lemma 4.1.
Proof of Theorem 2 (2). We have, with M = [δT1/3] for some small constant δ > 0, Z 2T T Gσ(t)3dt = Z 2T T Σ1,M(t)3dt + O( Z 2T T |ΣM,T|(Gσ(t)2+ Σ21,M(t)) dt). (4.2)
Proposition 4.1 and (4.1) yields that the O-term is O(T(13−8σ)/3). The integral on the right-sided of (4.2) is treated by the argument in [23]. Then the result follows.
5. Limiting distribution functions
We first quote some results from [1, Theorem 4.1] and [3, Theorem 6]. Let F be a real-valued function defined on [1, ∞), and let a1(t), a2(t), . . .
be real-valued, continuous and of period 1 such that R1
0 an(t) dt = 0 and
P∞
n=1
R1
γ2, . . . which are linearly independent over Q, such that lim N →∞lim supT →∞ 1 T Z T 0 min(1, |F (t) − X n≤N an(γnt)|) dt = 0. Fact I. For every continuous bounded function g on R, we have
lim T →∞ 1 T Z T 0 g(F (t)) dt = Z ∞ −∞ g(x)ν(dx), where ν(dx) is the distribution of the random series η =P∞
n=1an(tn) and tn
are independent random variables uniformly distributed on [0, 1]. Equiva-lently, the distribution function of F , PT(u) = T−1µ{t ∈ [1, T ] : F (t) ≤ u},
converges weakly to a function P (u), called the limiting distribution, as T → ∞.
Fact II. If RT
1 |F (t)|Adt T , then for any real k ∈ [0, A) and integral
l ∈ [0, A), the following limits exist: lim T →∞T −1Z T 1 |F (t)|kdt and lim T →∞T −1Z T 1 F (t)ldt. Now, let us take F (t) = t2σ−5/2G
σ(2πt2), γn= 2 √ n and (5.1) an(t) = √ 2 µ(n) 2 n7/4−σ ∞ X r=1 (−1)nrσ1−2σ(nr 2) r7/2−2σ sin(2πrt − π/4).
Following the computation in [3, p.402] with Lemma 4.1, we get
Z 2T T (F (t) − X n≤N an(2 √ nt))2dt T N2σ−5/2 (N ≤√T ).
Then Theorem 2 (c) and the first part of Theorem 3 are immediate conse-quence of Facts I and II with Proposition 4.1.
We proceed to prove the lower bounds in (2.2) with the idea in [1, Section V].
Lemma 5.1. Let n be squarefree. Define
An= {t ∈ [0, 1] : an(t) > B−1σ1−2σ(n)nσ−7/4} where B = 4A(P∞ r=1r4σ−7)−1 and A = √ 2P∞ r=1σ1−2σ(r2)r2σ−7/2. Then, we have µ(An) ≥ 1/(AB) where µ is the Lebesgue measure.
The proof makes use of the fact that R1
0 a+n(t) dt =
R1
0 a −
n(t) dt where
a±n(t) = max(0, ±an(t)), and
Z 1 0 a+n(t)2dt + Z 1 0 a−n(t)2dt = 1 n7/2−2σ ∞ X r=1 σ1−2σ(nr2)2 r7−4σ .
Proof of lower bounds in Theorem 3. By Markov’s inequality, we have Pr(| ∞ X m=n+1 am(tm)| ≤ 2 √ K) ≥ 1 − 1 4K ∞ X m=1 Z 1 0 am(t)2dt ≥ 3 4 where Pr(#) denotes the probability of the event # and
K = ∞ X m=1 Z 1 0 am(t)2dt < +∞.
Consider the set En= n (t1, t2, . . .) : tm∈ Am for 1 ≤ m ≤ n and | ∞ X m=n+1 am(tm)| ≤ 2 √ Ko where Am = [0, 1] if m is not squarefree. Then,
Pr(En) = n Y m=1 Pr(Am)Pr(| ∞ X m=n+1 am(tm)| ≤ 2 √ K) ≥ 3 4(AB)n
due to Pr(Am) = µ(Am) and Lemma 5.1. When (t1, t2, . . .) ∈ En, we have ∞ X m=1 am(tm) ≥ 1 B X m≤n m squarefree σ1−2σ(m) m7/4−σ − 2 √ K ( log n if σ = 3/4, nσ−3/4 if 3/4 < σ < 1.
Our result for 1 − Dσ(u) follows after we replace n by [eu] if σ = 3/4 and
by [u4/(4σ−3)] if 3/4 < σ < 1. The case of Dσ(−u) can be proved in the
same way.
To derive the upper estimates, we need a result on the Laplace transform of limiting distribution functions [13, Lemma 3.1].
Lemma 5.2. Let X be a real random variable with the probability distribu-tion D(x). Suppose D(x) > 0 for any x > 0. For the two cases: (i) ψ(x) =
x log x and φ(x) = log x, or (ii) ψ(x) = x4/(7−4σ) and φ(x) = x(4σ−3)/4, there exist two positive numbers L and L0 such that
(a) if lim supλ→∞ψ(λ)−1log E(exp(λX)) ≤ L, then lim sup
x→∞
x−1log(1 − D(φ(x))) ≤ −L0, (b) if lim supλ→∞ψ(λ)−1log E(exp(−λX)) ≤ L, then
lim sup
x→∞ x
Proof of upper bounds in Theorem 3. We take N = λ if σ = 3/4, and N = λ4/(7−4σ) if 3/4 < σ < 1. When n ≤ N , we use Z 1 0 exp(±λan(t)) dt ≤ exp(λA σ1−2σ(n)µ(n)2 n7/4−σ ). Recall that A =√2P∞
r=1σ1−2σ(r2)r7/2−2σ. Now consider n > N . If λAσ1−2σ(n) <
n7/4−σ, then by the inequality ex≤ 1+x+x2for x ≤ 1, andR1
0 an(t) dt = 0, we have Z 1 0 exp(±λan(t)) dt ≤ exp((λA)2 σ1−2σ(n)2µ(n)2 n7/2−2σ ).
Otherwise, λAσ1−2σ(n) ≥ n7/4−σ, it is obvious that
Z 1 0 exp(±λan(t)) dt ≤ exp(λA σ1−2σ(n)µ(n)2 n7/4−σ ) ≤ exp((λA)2σ1−2σ(n) 2µ(n)2 n7/2−2σ ).
Therefore, log E(exp(±λX)) is ≤ λA X n≤N σ1−2σ(n) n7/4−σ + (λA) 2 X n>N σ1−2σ(n)2 n7/2−2σ ( λ log λ if σ = 3/4, λ4/(7−4σ) if 3/4 < σ < 1. The proof is complete with Lemma 5.2.
6. Ω±-results
This section is devoted to prove Theorem 4. We apply the methods in [2] or [7], but beforehand, we transform Gσ(t) into a simple finite series by
convolution with the kernel
K(u) = 2B
sin 2πBu
2πBu
2
. Similarly to [15], we have, for 1 B L1/4 T1/16,
(6.1) t2σ−5/2 Z L −LGσ(2π(t + u) 2)K(u) du = S B(t) + O(B4σ−5) where (6.2) SB(t) = √ 2 X n≤B2 (−1)n(1 − √ n B ) σ1−2σ(n) n7/4−σ sin(4π √ nt −π 4). To prove the Ω−-result, we use Dirichlet’s Theorem to align the angles.
(1 + δ−B2)T1/10] such that kl√nk < δ. Taking B δ√log T , we have l ∈ [T1/10, T1/5] and SB(l) = − X n≤B2 (−1)n(1 − √ n B ) σ1−2σ(n) n7/4−σ + Oδ X n≤B2 (1 − √ n B ) σ1−2σ(n) n7/4−σ . (6.3)
A simple calculation shows that 22σ· X n≤B2 (−1)n(1 − √ n B ) σ1−2σ(n) n7/4−σ = X n≤B2 (1 − √ n B ) σ1−2σ(n) n7/4−σ .
We thus infer SB(t) = Ω−(log log t) if σ = 3/4, and Ω−((log t)σ−3/4) if
3/4 < σ < 1.
We proceed to prove the Ω+-result with the method in [2]. Take x =
δ log log T log log log T and B = T1/100(L = B4) for a small number δ > 0. We consider the convolution of SB(t) with a kernel involving the function
Tx(u) = Y q∈Qx (1 + cos(4π√qu)) = Y q∈Qx 1 +e 4πi√qu+ e−4πi√qu 2 !
where Qx is the set of positive squarefree integers whose prime factors
are odd and smaller than x. The convolution will pick out terms with the desired frequencies, Z ∞ −∞ SB(t + u)Tx(u) sin πu πu 2 du =√2 X n≤B2 n∈Qx (−1)n(1 − √ n B ) σ1−2σ(n) n7/4−σ sin(4π √ nt − π 4).
To maximize the right-hand side, we apply Dirichlet’s theorem again to find a number l ∈ [T1/10, (1 + δ−|Qx|)T1/10] so that the right-side is
Y p∈Qx 1 +σ1−2σ(p) p7/4−σ ( log x if σ = 3/4, exp(cxσ−3/4/(log x)) if σ > 3/4. This follows from the the estimation of P
p≤xpσ−7/4 for σ = 3/4 and
σ > 3/4 respectively. The cardinality of Qx is O(exp(cx/ log x)) for some
power of T . Consequently, we obtain sup T1/10uT1/4 SB(u) (
log log log T if σ = 3/4,
exp(c(log log T )σ−3/4(log log log T )σ−7/4) if σ > 3/4.
7. Occurrence of large values
Proof of Theorem 5. Define Kτ(u) = (1−|u|)(1+τ sin(4παu)) where τ = −1 or +1 and α is a large constant. Following the argument in [4], we derive that Z 1 −1(t + u) 2σ−5/2Z L −LGσ(2π(t + u + v) 2)K(v) dvK τ(u) du = τ 2(1 − B −1
) cos(4πt − π/4) + O(α−2) + O(B4σ−5). where δ1,n= 1 if n = 1 and 0 otherwise. Our assertion follows by choosing
B and α (L = B4) sufficiently large, and k4tk ≤ 1/8 with t ∈ [√T ,√T + 1].
(Note that τ can be +1 or −1 at our disposal.)
To prove Theorem 6, we need the next lemma which is the key.
Lemma 7.1. For T5/12≤ H ≤ T1/2, Z 2T T max 0≤h≤H(Gσ(t + h) − Gσ(t)) 2dt T H5−4σ where the implied constant depends on σ.
Proof. Following the arguments in [8], we have
(7.1)
Z 2T
T
(Gσ(t + h) − Gσ(t))2dt T h5−4σmin((σ − 3/4)−1, log(T /h2))
where log2T ≤ h ≤√T . Let b = T1/24 and H = 2λb. Then, as in [4], we can show
max
0≤h≤H|Gσ(t + h) − Gσ(t)| ≤ max1≤j≤2λ|Gσ(t + jb) − Gσ(t)| + O(T
2(1−σ)/3+b
for any fixed t. Let us take 1 ≤ j0= j0(t) ≤ 2λ such that
|Gσ(t + j0b) − Gσ(t)| = max
1≤j≤2λ|Gσ(t + jb) − Gσ(t)|.
Then we can express j0 = 2λPµ∈St2−µ for some set St of non-negative
integers. Hence,
Gσ(t + j0b) − Gσ(t) =
X
µ∈St
where 0 ≤ ν = νt,µ< 2µis an integer. By Cauchy-Schwarz’s inequality and
inserting the remaining ν’s, we get (Gσ(t + j0b) − Gσ(t))2 ≤ X µ∈St 2−(1−σ)µ X µ∈St 2(1−σ)µ(Gσ(t + (ν + 1)2λ−µb) − Gσ(t + ν2λ−µb))2 X µ∈St X 0≤ν<2µ 2(1−σ)µ(Gσ(t + (ν + 1)2λ−µb) − Gσ(t + ν2λ−µb))2 asP
µ∈St2−(1−σ)µ 1. Integrating over [T, 2T ] and using (7.1), we see that
Z 2T T max 0≤h≤H(Gσ(t + h) − Gσ(t)) 2dt X µ∈St X 0≤ν<2µ 2(1−σ)µ Z 2T +ν2λ−µb T +ν2λ−µb (Gσ(t + 2 λ−µb) − G σ(t))2dt + T17/12+ T H5−4σ X µ∈St X 0≤ν<2µ 2−(4−3σ)µ T H5−4σ.
This complete the proof of Lemma 7.1.
Proof of Theorem 6. Define G±σ(t) = max(0, ±Gσ(t)). By Theorem 2 (c),
we have R2T
T |Gσ(t)|3dt T1+3(5/4−σ). Hence, Cauchy-Schwarz inequality
gives Z 2T T Gσ(t)2dt !2 ≤ Z 2T T |Gσ(t)| dt Z 2T T |Gσ(t)|3dt. we have R2T T |Gσ(t)| dt T1+(5/4−σ). Together with (2.1),RT2TG±σ(t) dt ≥ c12RT2Tt5/4−σdt.
Consider K±(t) = G±σ(t) − (c12− )t5/4−σ where = δ5/2−2σ, we have
Z 2T T K±(t) dt ≥ Z 2T T t5/4−σdt
and K±(t + h) − K±(t) = G±σ(t + h) − G±σ(t) + O(T1/4−σh). Since |G±σ(t + h) − G±σ(t)| ≤ |Gσ(t + h) − Gσ(t)|, it follows that together with
Lemma 7.1, Z 2T T max h≤H|K ±(t + h) − K±(t)| dt T H5/2−2σ+ T5/4−σH.
Define ω±(t) = K±(t) − maxh≤H|K±(t + h) − K±(t)|. Taking H =
have Z 2T T ω±(t) dt ≥ Z 2T T t5/4−σdt − Z 2T T max h≤H|K ± (t + h) − K±(t)| dt T1+(5/4−σ). Let I±= {t ∈ [T, 2T ] : ω±(t) > 0}. Then Z 2T T ω±(t) dt ≤ Z I±ω ±(t) dt ≤Z I±K ±(t) dt ≤ Z I± dt 1/2 Z 2T T K±(t)2dt !1/2 . We infer |I±| 2T asR2T T K±(t)2dt R2T T Gσ(t)2dt+T7/2−2σ. When t ∈
I±, we have K±(t) ≥ maxh≤H|K±(t + h) − K±(t)| ≥ 0. Hence, K±(u) ≥ 0
for all u ∈ [t, t + H], i.e. G±σ(t) ≥ (c12− )t5/4−σ. The number of such
intervals is not less than |I±|/H c13δ4(1−σ)
√ T .
Acknowledgement
The author would thank the referee for the valuable comments. Also, he wishes to express his sincere gratitude to Professor Kohji Matsumoto for his warm encouragement. Hearty thanks are due to Charlies Tu for unfailing supports.
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Yuk-Kam Lau
Department of Mathematics The University of Hong Kong Pokfulam Road, Hong Kong